$\frac{{0.693}}{{{{\text{k}}_1}}} = \frac{{{{\text{C}}_0}}}{{2{{\text{k}}_0}}}$ $ \Rightarrow \frac{{{{\text{k}}_1}}}{{{{\text{k}}_0}}} = \frac{{2 \times 0.693}}{{{{\text{C}}_0}}}$
$\mathrm{r}_{1}=\mathrm{k}_{1} \times\left[\mathrm{C}_{0}\right]^{1}$
$\mathrm{r}_{0}=\mathrm{k}_{0} \times\left[\mathrm{C}_{0}\right]^{0}$
$\frac{\mathrm{r}_{1}}{\mathrm{r}_{0}}=\frac{\mathrm{k}_{1}\left[\mathrm{C}_{0}\right]}{\mathrm{k}_{0}}=2 \times 0.693$
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$E_{{A^{3 + }}/A}^o = 1.50\,\,V\,,$ $E_{{B^{2 + }}/B}^o = 0.3\,\,V,$
$E_{{C^{3 + }}/C}^o = - \,0.74\,\,V,$ $E_{{D^{2 + }}/D}^o = - \,2.37\,\,V.$
The correct sequence in which the various metals are deposited at the cathode is
